# NCERT Solutions for Class 9 Maths Chapter 10 Circles

NCERT solutions for class 9 maths chapter 10 Circles:  A round figure made up using several points that are equidistant from a reference point is called a circle. You can relate this chapter with real-life things. Many objects that we come across in our daily life is circular in shape, such as ring, bangle, wheels of the vehicle, clock, etc. CBSE NCERT solutions for class 9 maths chapter 10 Circles will help in solving all problems related to circular shapes. The circle is an integral part of unit geometry. A circle divides the plane in which the circle lies into three parts as shown in figure 1, which are the interior of the circle, the exterior of the circle and the circle. In this particular chapter, you will learn the proof of theorems and problems based on those theorems. Solutions of NCERT for class 9 maths chapter 10 Circles are designed in such a way that a student can fetch 100% marks in a particular question. In the upcoming class, you will get to know some advanced versions of circles but the advanced part can only be done when you have done the basics. So, class 9 maths circles are a foundation for next class board examinations. NCERT solutions for class 9 maths chapter 10 circles will assist you in building this strong foundation. NCERT solutions are also available for different chapters, classes, and subjects which you can get by clicking on the given link.

## NCERT solutions for class 9 maths chapter 10 Circles Excercise: 10.1

Fill in the blanks:

The centre of a circle lies in the interior of the circle.

Fill in the blanks:

A point, whose distance from the centre of a circle is greater than its radius lies in exterior of the circle.

Fill in the blanks:

The longest chord of a circle is a  diameter of the circle.

Fill in the blanks:

An arc is a semi- circle when its ends are the ends of a diameter.

Fill in the blanks:

Segment of a circle is the region between an arc and chord of the circle.

Fill in the blanks:

A circle divides the plane, on which it lies, in two parts.

True. As line segment joining the centre to any point on the circle is a radius of the circle.

False . As a circle has infinite number of equal chords.

False.  If a circle is divided into three equal arcs, each arc makes angle of 120 degrees whereas major arc makes angle greater than 180 degree at centre.

True.A chord of a circle, which is twice as long as its radius, is a diameter of the circle.

False. As the sector is the region between the radii and arc.

Q2 (vi) A circle is a plane figure.

## Q1 Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres.

Given: The two circles are congruent if they have the same radii.

To prove: The equal chords of congruent circles subtend equal angles at their centres i.e. $\angle$BAC=$\angle$QPR

Proof :

In $\triangle$ABC and $\triangle$PQR,

BC = QR         (Given)

AB = PQ          (Radii of congruent circle)

AC = PR          (Radii of congruent circle)

Thus,  $\triangle$ABC $\cong$ $\triangle$PQR         (By SSS rule)

$\angle$BAC=$\angle$QPR        (CPCT)

Given : chords of congruent circles subtend equal angles at their centres,

To prove :  BC = QR

Proof :

In $\triangle$ABC and $\triangle$PQR,

$\angle$BAC=$\angle$QPR       (Given)

AB = PQ          (Radii of congruent circle)

AC = PR          (Radii of congruent circle)

Thus,  $\triangle$ABC $\cong$ $\triangle$PQR         (By SAS rule)

BC = QR             (CPCT)

## NCERT solutions for class 9 maths chapter 10 Circles Excercise: 10.3

In (i) we do not have any common point.

In (ii) we have 1 common point.

In (iii) we have 1 common point.

In (iv) we have 2 common points.

The maximum number of common points is 2.

Given : Points P,Q,R lies on circle.

Construction :

1. Join PR and QR

2. Draw perpendicular bisector of PR and QR which intersects at point O.

3. Taking O as centre and OP as radius draw a circle.

4. The circle obtained is required.

Given: Two circles intersect at two points.

To prove: their centres lie on the perpendicular bisector of the common chord.

Construction: Joinpoint P and Q to midpoint M of chord AB.

Proof: AB is a chord of circle C(Q,r)  and QM is the bisector of chord AB.

$\therefore PM\perp AB$

Hence, $\angle PMA =90 \degree$

Similarly, AB is a chord of circle(Q,r' ) and QM is the bisector of chord AB.

$\therefore QM\perp AB$

Hence, $\angle QMA =90 \degree$

Now, $\angle QMA +\angle PMA=90 \degree+90 \degree= 180 \degree$

$\angle$PMA and $\angle$QMA are forming linear pairs so PMQ is a straight line.

Hence, P and Q lie on the perpendicular bisector of common chord AB.

## NCERT solutions for class 9 maths chapter 10 Circles Excercise: 10.4

Given: Two circles of radii $\small 5\hspace{1mm}cm$ and $\small 3\hspace{1mm}cm$ intersect at two points and the distance between their centres is $\small 4\hspace{1mm}cm$.

To find the length of the common chord.

Construction: Join OP and draw $OM\perp AB\, \, and \, \, \, ON\perp CD.$

Proof: AB is a chord of circle C(P,3)  and PM is the bisector of chord AB.

$\therefore PM\perp AB$

$\angle PMA=90 \degree$

Let, PM = x , so QM=4-x

In $\triangle$APM, using Pythagoras theorem

$AM^2=AP^2-PM^2$...........................1

Also,

In $\triangle$AQM, using Pythagoras theorem

$AM^2=AQ^2-MQ^2$...........................2

From 1 and 2, we get

$AP^2-PM^2=AQ^2-MQ^2$

$\Rightarrow 3^2-x^2=5^2-(4-x)^2$

$\Rightarrow 9-x^2=25-16-x^2+8x$

$\Rightarrow 9=9+8x$

$\Rightarrow 8x=0$

$\Rightarrow x=0$

Put,x=0 in equation 1

$AM^2=3^2-0^2=9$

$\Rightarrow AM=3$

$\Rightarrow AB=2AM=6$

Given: two equal chords of a circle intersect within the circle

To prove: Segments of one chord are equal to corresponding segments of the other chord i.e. AP = CP and BP=DP.

Construction : Join OP and draw $OM\perp AB\, \, \, \, and\, \, \, ON\perp CD.$

Proof :

In $\triangle$OMP and $\triangle$ONP,

AP = AP         (Common)

OM = ON          (Equal chords of a circle are equidistant from the centre)

$\angle$OMP = $\angle$ONP      (Both are right angled)

Thus,  $\triangle$OMP $\cong$ $\triangle$ONP         (By SAS rule)

PM = PN..........................1 (CPCT)

AB = CD ............................2(Given )

$\Rightarrow \frac{1}{2}AB=\frac{1}{2}CD$

$\Rightarrow AM = CN$......................3

Adding 1 and 3, we have

AM + PM = CN + PN

$\Rightarrow AP = CP$

Subtract 4 from 2, we get

AB-AP = CD - CP

$\Rightarrow PB = PD$

Given: two equal chords of a circle intersect within the circle.

To prove: the line joining the point of intersection to the centre makes equal angles with the chords.
i.e. $\angle$OPM=$\angle$OPN

Proof :

Construction: Join OP and draw $OM\perp AB\, \, \, \, and\, \, \, ON\perp CD.$

In $\triangle$OMP and $\triangle$ONP,

AP = AP         (Common)

OM = ON          (Equal chords of a circle are equidistant from the centre)

$\angle$OMP = $\angle$ONP      (Both are right-angled)

Thus,  $\triangle$OMP $\cong$ $\triangle$ONP         (By RHS rule)

$\angle$OPM=$\angle$OPN   (CPCT)

Given: a line intersects two concentric circles (circles with the same centre) with centre O at A, B,  C and D.

To prove : AB  = CD

Construction: Draw $OM\perp AD$

Proof :

BC is a chord of the inner circle and $OM\perp BC$

So, BM = CM  .................1

(Perpendicular OM bisect BC)

Similarly,

AD  is a chord of the outer circle and $OM\perp AD$

So, AM = DM  .................2

Subtracting   1 from  2, we get

AM-BM = DM - CM

$\Rightarrow AB = CD$

Given: From the figure, R, S, M are the position of Reshma, Salma, Mandip respectively.

So, RS = SM = 6 cm

Construction : Join OR,OS,RS,RM and OM.Draw $OL\perp RS$.

Proof:

In $\triangle$ ORS,

OS = OR    and  $OL\perp RS$ (by construction )

So,       RL = LS = 3cm        (RS = 6 cm )

In $\triangle$OLS, by pytagoras theorem,

$OL^2=OS^2-SL^2$

$\Rightarrow OL^2=5^2-3^2=25-9=16$

$\Rightarrow OL=4$

In   $\triangle$ ORK and $\triangle$OMK,

$\angle$ROK = $\angle$ MOK    (Equal chords subtend equal angle at centre)

OK = OK         (Common)

$\triangle$ ORK $\cong$ $\triangle$OMK    (By SAS)

RK = MK    (CPCT)

Thus,$OK\perp RM$

area of $\triangle$ORS  =$\frac{1}{2}\times RS\times OL$...............................1

area of $\triangle$ORS  =$\frac{1}{2}\times OS\times KR$.............................2

From 1 and 2, we get

$\frac{1}{2}\times RS\times OL$  $=\frac{1}{2}\times OS\times KR$

$\Rightarrow RS\times OL=OS\times KR$

$\Rightarrow 6\times 4=5\times KR$

$\Rightarrow KR=4.8 cm$

Thus, $RM =2 KR=2\times 4.8 cm=9.6 cm$

Given: In the figure, A, S, D are positioned Ankur, Syed and David respectively.

So, AS = SD = AD

Radius of circular park = 20 m

so, AO=SO=DO=20 m

Construction: AP$\perp$ SD

Proof :

Let AS = SD = AD = 2x cm

In $\triangle$ASD,

AS = AD   and     AP$\perp$ SD

So,  SP = PD = x cm

In $\triangle$OPD, by Pythagoras,

$OP^2=OD^2-PD^2$

$\Rightarrow OP^2=20^2-x^2=400-x^2$

$\Rightarrow OP=\sqrt{400-x^2}$

In $\triangle$APD, by Pythagoras,

$AP^2=AD^2-PD^2$

$\Rightarrow (AO+OP)^2+x^2=(2x)^2$

$\Rightarrow (20+\sqrt{400-x^2})^2+x^2=4x^2$

$\Rightarrow 400+400-x^2+40\sqrt{400-x^2}+x^2=4x^2$

$\Rightarrow 800+40\sqrt{400-x^2}=4x^2$

$\Rightarrow 200+10\sqrt{400-x^2}=x^2$

$\Rightarrow 10\sqrt{400-x^2}=x^2-200$

Squaring both sides,

$\Rightarrow 100(400-x^2)=(x^2-200)^2$

$\Rightarrow 40000-100x^2=x^4-40000-400x^2$

$\Rightarrow x^4-300x^2=0$

$\Rightarrow x^2(x^2-300)=0$

$\Rightarrow x^2=300$

$\Rightarrow x=10\sqrt{3}$

Hence, length of string of each phone$= 2x=20\sqrt{3}$m

## NCERT solutions for class 9 maths chapter 10 Circles Excercise: 10.5

$\angle$AOC = $\angle$AOB + $\angle$BOC= $60 \degree+30 \degree=90 \degree$

$\angle$AOC = 2 $\angle$ADC   (angle subtended by an arc at the centre is double the angle subtended by it at any)

$\angle ADC=\frac{1}{2}\angle AOC$

$\Rightarrow \angle ADC=\frac{1}{2}90 \degree= 45 \degree$

Given: A chord of a circle is equal to the radius of the circle i.e. OA=OB.

To find: ADB and $\angle$ACB.

Solution :

In $\triangle$OAB,

OA = AB        (Given )

OA = OB          (Radii of circle)

So,  OA=OB=AB

$\Rightarrow$ ABC is a equilateral triangle.

So,$\angle$AOB = $60 \degree$

$\angle$AOB  = 2 $\angle$ADB

$\Rightarrow \angle ADB=\frac{1}{2}\angle AOB$

$\Rightarrow \angle ADB=\frac{1}{2}60 \degree=30$

ACBD is a cyclic quadrilateral .

So, $\angle$ACB+$\angle$ADB =$180 \degree$

$\Rightarrow \angle ACB+30 \degree= 180 \degree$

$\Rightarrow \angle ACB= 180 \degree-30 \degree=150 \degree$

Construction: Join PS and RS.

So, $\angle$PSR + $\angle$PQR = $180 \degree$

$\Rightarrow \angle PSR+100 \degree=180 \degree$

$\Rightarrow \angle PSR=180 \degree-100 \degree=80 \degree$

Here, $\angle$POR = 2 $\angle$PSR

$\Rightarrow \angle POR=2\times 80 \degree=160 \degree$

In $\triangle$OPR ,

$\angle$ORP = $\angle$OPR    (the angles opposite to equal sides)

In $\triangle$OPR ,

$\angle$OPR+$\angle$ORP+$\angle$POR=$180 \degree$

$\Rightarrow 2\angle OPR+160 \degree= 180 \degree$

$\Rightarrow 2\angle OPR=180 \degree- 160 \degree$

$\Rightarrow 2\angle OPR=20 \degree$

$\Rightarrow \angle OPR=10 \degree$

In $\triangle$ ABC,

$\angle$A+$\angle$ABC+$\angle$ACB=$180\degree$

$\Rightarrow \angle A+69 \degree+31 \degree=180\degree$

$\Rightarrow \angle A+100 \degree=180\degree$

$\Rightarrow \angle A=180 \degree-100\degree$

$\Rightarrow \angle A=80 \degree$

$\angle$A  = $\angle$BDC = $80 \degree$   (Angles in same segment)

.

$\angle$DEC+ $\angle$BEC = $180 \degree$          (linear pairs)

$\Rightarrow$ $\angle$DEC+ $130 \degree$ = $180 \degree$           ($\angle$ BEC =$130 \degree$)

$\Rightarrow$ $\angle$DEC = $180 \degree$ - $130 \degree$

$\Rightarrow$ $\angle$DEC = $50 \degree$

In $\triangle$ DEC,

$\angle$D+ $\angle$DEC+ $\angle$DCE = $180 \degree$

$\Rightarrow \angle D+50 \degree+20 \degree= 180 \degree$

$\Rightarrow \angle D+70 \degree= 180 \degree$

$\Rightarrow \angle D= 180 \degree-70 \degree=110 \degree$

$\angle$D = $\angle$BAC     (angles in same segment are equal  )

$\angle$BAC  = $110 \degree$

$\angle BDC=\angle BAC$       (angles in the same segment are equal )

$\angle BDC= 30 \degree$

In $\triangle BDC,$

$\angle BCD+\angle BDC+\angle DBC= 180 \degree$

$\Rightarrow \angle BCD+30 \degree+70 \degree= 180 \degree$

$\Rightarrow \angle BCD+100 \degree= 180 \degree$

$\Rightarrow \angle BCD=180 \degree- 100 \degree=80 \degree$

If  AB = BC ,then

$\angle BCA=\angle BAC$

$\Rightarrow \angle BCA=30 \degree$

Here, $\angle ECD+\angle BCE=\angle BCD$

$\Rightarrow \angle ECD+30 \degree=80 \degree$

$\Rightarrow \angle ECD=80 \degree-30 \degree=50 \degree$

AC is the diameter of the circle.

Thus,$\angle ADC=90 \degree$   and    $\angle ABC=90 \degree$............................1(Angle in a semi-circle is right angle)

Similarly, BD is the diameter of the circle.

Thus,$\angle BAD=90 \degree$   and    $\angle BCD=90 \degree$............................2(Angle in a semi-circle is right angle)

From 1 and 2, we get

$\angle BCD=\angle ADC=\angle ABC=\angle BAD =90 \degree$

Hence, ABCD is a rectangle.

Given: ABCD is a trapezium.

AB || DE    (Given )

AD || BE    ( By construction )

Thus, ABED is a parallelogram.

AD = BE     (Opposite sides of parallelogram )

so,  BE = BC

In $\triangle$EBC,

BE = BC    (Proved above )

Thus, $\angle C = \angle 2$...........1(angles opposite to equal sides )

$\angle A= \angle 1$...............2(Opposite angles of the parallelogram )

From 1 and 2, we get

$\angle 1+\angle 2=180 \degree$           (linear pair)

$\Rightarrow \angle A+\angle C=180 \degree$

Thus, ABED is a cyclic quadrilateral.

$\angle ABP=\angle QBD$................1(vertically opposite angles)

$\angle ACP=\angle ABP$..................2(Angles in the same segment are equal)

$\angle QBD=\angle QCD$.................3(angles in the same segment are equal)

From 1,2,3 ,we get

$\angle ACP=\angle QCD$

Given: circles are drawn taking two sides of a triangle as diameters.

Proof: AB is the diameter of the circle and $\angle$ADB is formed in a semi-circle.

$\angle$ADB = $90 \degree$........................1(angle in a semi-circle)

Similarly,

AC is the diameter of the circle and $\angle$ADC is formed in a semi-circle.

$\angle$ADC = $90 \degree$........................2(angle in a semi-circle)

From 1 and 2, we have

$\angle$ADB+$\angle$ADC=$90 \degree$+$90 \degree$=$180 \degree$

$\angle$ADB and $\angle$ADC are forming a linear pair. So, BDC is a straight line.

Hence, point D lies on this side.

Given: ABC and ADC are two right triangles with common hypotenuse AC.

To prove : $\small \angle CAD =\angle CBD$

Proof :

Triangle ABC and ADC are on common base BC and $\angle$BAC = $\angle$BDC.

Thus, point A,B,C,D lie in the same circle.

(If a line segment joining two points subtends equal angles at two other points lying on the same side of line containing line segment, four points lie on the circle.)

$\angle$CAD = $\angle$CBD  (Angles in same segment are equal)

Given: ABCD is a cyclic quadrilateral.

To prove: ABCD is a rectangle.

Proof :

$\angle A + \angle C = 180 \degree$.......................1(sum of either pair of opposite angles of a cyclic quadrilateral)

$\angle A = \angle C$........................................2(opposite angles of a parallelogram are equal )

From 1 and 2,

$\angle A + \angle A = 180 \degree$

$\Rightarrow 2\angle A = 180 \degree$

$\Rightarrow \angle A = 90 \degree$

We know that a parallelogram with one angle right angle is a rectangle.

Hence, ABCD is a rectangle.

## NCERT solutions for class 9 maths chapter 10 Circles Excercise: 10.6

Given: Circle C(P,r) and circle C(Q,r') intersect each other at A and B.

To prove : $\angle$PAQ = $\angle$PBQ

Proof : In $\triangle$APQ and $\triangle$BPQ,

PA = PB         (radii of same circle)

PQ = PQ        (Common)

QA = QB       (radii of same circle)

So,           $\triangle$APQ $\cong$ $\triangle$BPQ       (By SSS)

$\angle$PAQ = $\angle$PBQ     (CPCT)

Given : AB = 5 cm, CD = 11 cm and AB || CD.

Construction: Draw $OM \perp CD \, \, and \, \, \, ON\perp AB$

Proof :

Proof: CD is a chord of circle  and  $OM \perp CD$

Thus, CM = MD = 5.5 cm    (perpendicular from centre bisects chord)

and      AN = NB  = 2.5 cm

Let OM be x.

So, ON = 6 - x              (MN = 6 cm )

In $\triangle$ OCM , using Pythagoras,

$OC ^2=CM^2+OM^2$.............................1

and

In $\triangle$ OAN , using Pythagoras,

$OA ^2=AN^2+ON^2$.............................2

From 1 and 2,

$CM ^2+OM^2=AN^2+ON^2$            (OC=OA =radii)

$5.5 ^2+x^2=2.5^2+(6-x)^2$

$\Rightarrow 30.25+x^2=6.25+36+x^2-12x$

$\Rightarrow 30.25-42.25=-12x$

$\Rightarrow -12=-12x$

$\Rightarrow x=1$

From 2, we get

$OC^2=5.5^2+1^2=30.25+1=31.25$

$\Rightarrow OC=\frac{5}{2}\sqrt{5} cm$

OA = OC

Thus, the radius of the circle is $\frac{5}{2}\sqrt{5} cm$

Given : AB = 8 cm, CD = 6 cm , OM = 4 cm and AB || CD.

To find: Length of ON

Construction: Draw $OM \perp CD \, \, and \, \, \, ON\perp AB$

Proof :

Proof: CD is a chord of circle  and  $OM \perp CD$

Thus, CM = MD = 3 cm    (perpendicular from centre bisects chord)

and      AN = NB  = 4 cm

Let MN be x.

So, ON = 4 - x              (MN = 4 cm )

In $\triangle$ OCM , using Pythagoras,

$OC ^2=CM^2+OM^2$.............................1

and

In $\triangle$ OAN , using Pythagoras,

$OA ^2=AN^2+ON^2$.............................2

From 1 and 2,

$CM ^2+OM^2=AN^2+ON^2$            (OC=OA =radii)

$\Rightarrow 3 ^2+4^2=4^2+(4-x)^2$

$\Rightarrow 9+16=16+16+x^2-8x$

$\Rightarrow 9=16+x^2-8x$

$\Rightarrow x^2-8x+7=0$

$\Rightarrow x^2-7x-x+7=0$

$\Rightarrow x(x-7)-1(x-7)=0$

$\Rightarrow (x-1)(x-7)=0$

$\Rightarrow x=1,7$

So, x=1  (since  $x\neq 7> OM$)

ON =4-x =4-1=3 cm

Hence, second chord is 3 cm away from centre.

To prove : $\angle ABC = \frac{1}{2}(\angle AOC-\angle DOE)$

Construction: Join AC and DE.

Proof :

Let $\angle$ADC = x , $\angle$DOE = y and  $\angle$AOD = z

So,  $\angle$EOC = z (each chord subtends equal angle at centre)

$\angle$AOC + $\angle$DOE +$\angle$AOD + $\angle$EOC = $360 \degree$

$\Rightarrow x+y+z+z=360 \degree$

$\Rightarrow x+y+2z=360 \degree$.........................................1

In $\triangle$ OAD ,

OA = OD  (Radii of the circle)

$\angle$OAD = $\angle$ODA    (angles opposite to equal sides )

$\angle$OAD + $\angle$ODA + $\angle$AOD =$180 \degree$

$\Rightarrow 2\angle OAD+z=180 \degree$

$\Rightarrow 2\angle OAD=180 \degree-z$

$\Rightarrow \angle OAD=\frac{180 \degree-z}{2}$

$\Rightarrow \angle OAD=90 \degree-\frac{z}{2}$.............................................................2

Similarly,

$\Rightarrow \angle OCE=90 \degree-\frac{x}{2}$.............................................................3

$\Rightarrow \angle OED=90 \degree-\frac{y}{2}$..............................................................4

$\angle$ODB is exterior of triangle OAD . So,

$\angle$ODB = $\angle$OAD + $\angle$ODA

$\Rightarrow \angle ODB=90 \degree-\frac{z}{2}+z$                  (from  2)

$\Rightarrow \angle ODB=90 \degree+\frac{z}{2}$.................................................................5

similarly,

$\angle$OBE is exterior of triangle OCE . So,

$\angle$OBE = $\angle$OCE + $\angle$OEC

$\Rightarrow \angle OEB=90 \degree-\frac{z}{2}+z$                  (from  3)

$\Rightarrow \angle OEB=90 \degree+\frac{z}{2}$.................................................................6

From 4,5,6 ;we get

$\angle$BDE = $\angle$BED = $\angle$OEB - $\angle$OED

$\Rightarrow \angle BDE=\angle BED=90 \degree+\frac{z}{2}-(90-\frac{y}{2})=\frac{y+z}{2}$

$\Rightarrow \angle BDE+\angle BED=y+z$..................................................7

In $\triangle$BDE ,

$\angle$DBE + $\angle$BDE + $\angle$BED = $180 \degree$

$\Rightarrow \angle DBE +y+z=180 \degree$

$\Rightarrow \angle DBE =180 \degree-(y+z)$

$\Rightarrow \angle ABC =180 \degree-(y+z)$...................................................8

Here,  from equation 1,

$\frac{x-y}{2}=\frac{360 \degree-y-2x-y}{2}$

$\Rightarrow \frac{x-y}{2}=\frac{360 \degree-2y-2x}{2}$

$\Rightarrow \frac{x-y}{2}=180 \degree-y-x$...................................9

From 8 and 9,we have

$\angle ABC=\frac{x-y}{2}=\frac{1}{2}(\angle AOC-\angle DOE)$

Given : ABCD is rhombus.

To prove : the circle drawn with AB  as diameter, passes through the  point O.

Proof :

ABCD is rhombus.

Thus, $\angle AOC = 90 \degree$             (diagonals of a rhombus bisect each other at $90 \degree$)

So, a circle drawn AB as diameter will pass through point O.

Thus,  the circle is drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.

Given: ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E.

To prove :   AE = AD

Proof :

$\angle$ADC = $\angle$3  , $\angle$ABC = $\angle$4, $\angle$ADE = $\angle$1  and $\angle$AED = $\angle$2

$\angle 3+\angle 1=180 \degree$.................1(linear pair)

$\angle 2+\angle 4=180 \degree$....................2(sum of opposite angles of cyclic quadrilateral)

$\angle$3 = $\angle$4      (oppsoite angles of parallelogram )

From 1 and 2,

$\angle$3+$\angle$1 = $\angle$2 + $\angle$

From 3,    $\angle$1 = $\angle$

From 4,    $\triangle$AQB,     $\angle$1 = $\angle$2

Therefore, AE = AD (In an isosceles triangle ,angles oppsoite to equal sides are equal)

Given: AC and BD are chords of a circle which bisect each other.

To prove: AC and BD are diameters.

Construction : Join AB,BC,CD,DA.

Proof :

In $\triangle$ ABD and $\triangle$CDO,

AO = OC     (Given )

$\angle$AOB = $\angle$COD  (Vertically opposite angles )

BO = DO        (Given )

So,  $\triangle$ ABD$\cong$ $\triangle$CDO     (By SAS)

$\angle$BAO = $\angle$DCO    (CPCT)

$\angle$BAO and $\angle$DCO are alternate angle and are equal .

So,     AB || DC ..............1

From 1 and 2,

$\angle A+\angle C=180 \degree$......................3(sum of opposite angles)

$\angle$A = $\angle$C    ................................4(Opposite angles of the parallelogram )

From 3 and 4,

$\angle A+\angle A=180 \degree$

$\Rightarrow 2\angle A=180 \degree$

$\Rightarrow \angle A=90 \degree$

BD is a diameter of the circle.

Similarly, AC is a diameter.

Given:  AC and BD are chords of a circle which bisect each other.

To prove: ABCD is a rectangle.

Construction : Join AB,BC,CD,DA.

Proof :

ABCD is a parallelogram. (proved in (i))

$\angle A=90 \degree$                    (proved in (i))

A parallelogram with one angle $90 \degree$, is a rectangle )

Thus, ABCD is rectangle.

Given :   Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively.

To prove :  the angles of the triangle DEF are    $\small 90^{\circ}-\frac{1}{2}C$,    $\small 90^{\circ}-\frac{1}{2}B$ and $\small 90^{\circ}-\frac{1}{2}A$

Proof :

$\angle$1 and $\angle$3 are angles in same segment.therefore,

$\angle$1 = $\angle$3 ................1(angles in same segment are equal )

and    $\angle$2 = $\angle$4 ..................2

$\angle$1+$\angle$2=$\angle$3+$\angle$4

$\Rightarrow \angle D=\frac{1}{2}\angle B+\frac{1}{2}\angle C$,

$\Rightarrow \angle D=\frac{1}{2}(\angle B+\angle C)$

$\Rightarrow \angle D=\frac{1}{2}(180 \degree+\angle C)$

and  $\Rightarrow \angle D=\frac{1}{2}(180 \degree-\angle A)$

$\Rightarrow \angle D=90 \degree-\frac{1}{2}\angle A$

Similarly,  $\Rightarrow \angle E=90 \degree-\frac{1}{2}\angle B$     and    $\angle F=90 \degree-\frac{1}{2}\angle C$

Given: Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles.

To prove :  BP = BQ

Proof :

AB is a common chord in both congruent circles.

$\therefore \angle APB = \angle AQB$

In $\triangle BPQ,$

$\angle APB = \angle AQB$

$\therefore BQ = BP$      (Sides opposite to equal of the triangle are equal )

Given :In any triangle ABC, if the angle bisector of  $\small \angle A$ and perpendicular bisector of BC  intersect.

To prove : D lies on perpendicular bisector BC.

Construction: Join BD and DC.

Proof :

Let $\angle$ABD = $\angle$1 ,  $\angle$ADC = $\angle$2 , $\angle$DCB =$\angle$3 , $\angle$CBD = $\angle$4

$\angle$1 and $\angle$3 lies in same segment.So,

$\angle$1 = $\angle$3    ..........................1(angles in same segment)

similarly, $\angle$2 = $\angle$4  ......................2

also,       $\angle$1=$\angle$2 ..............3(given)

From 1,2,3 , we get

$\angle$3 = $\angle$4

Hence,  BD = DC (angles opposite to equal sides are equal )

All points lying on perpendicular bisector BC will be equidistant from B and C.

Thus, point D also lies on perpendicular bisector BC.

## NCERT solutions for class 9 maths chapter wise

 Chapter No. Chapter Name Chapter 1 NCERT solutions for class 9 maths chapter 1 Number Systems Chapter 2 CBSE NCERT solutions for class 9 maths chapter 2 Polynomials Chapter 3 Solutions of NCERT class 9 maths chapter 3 Coordinate Geometry Chapter 4 NCERT solutions for class 9 maths chapter 4 Linear Equations In Two Variables Chapter 5 CBSE NCERT solutions for class 9 maths chapter 5 Introduction to Euclid's Geometry Chapter 6 Solutions of NCERT class 9 maths chapter 6 Lines And Angles Chapter 7 NCERT solutions for class 9 maths chapter 7 Triangles Chapter 8 CBSE NCERT solutions for class 9 maths chapter 8 Quadrilaterals Chapter 9 Solutions of NCERT class 9 maths chapter 9 Areas of Parallelograms and Triangles Chapter 10 NCERT solutions for class 9 maths chapter 10 Circles Chapter 11 CBSE NCERT solutions for class 9 maths chapter 11 Constructions Chapter 12 Solutions of NCERT class 9 maths chapter 12 Heron’s Formula Chapter 13 NCERT solutions for class 9 maths chapter 13 Surface Area and Volumes Chapter 14 CBSE NCERT solutions for class 9 maths chapter 14 Statistics Chapter 15 Solutions of NCERT class 9 maths chapter 15 Probability

## How to use NCERT solutions for class 9 maths chapter 10 Circles

• Learn and memorize some theorems related to circles

• Learn the application of those theorems in the problems

• Start applying the theorems and concepts on the practice exercises.

• During the practice exercises, you can take the help of NCERT solutions for class 9 maths chapter 10 Circles.

• After doing all the above-written things, you can practice more using the previous year questions papers.

Keep working hard & happy learning!